![[Maple Plot]](images/2-71.gif)
Gradients and Level Curves
If the graph of a function is sliced with a horizontal plane, the resulting curve of intersection is called a level curve of the function. In particular, if g ( x,y ) is a function, then the curve g ( x,y ) = k in the xy -plane is the curve formed by the intersection of the surface z = g ( x,y ) and the horizontal plane z=k.
(commands used to generate the figure above)
To study such curves, we use the gradient of the function. In particular, the gradient of g ( x,y ), which is sometimes denoted by grad(g), is defined
grad(g) =
![<g[x],g[y]>](images/2-72.gif){kind=small}
Moreover, it has the property that it is perpendicular to the tangent plane at a given point. That is, if m is a tangent vector to the curve at a given point ( p,q ), then the gradient at ( p,q ) is orthogonal to m.
![[Maple Plot]](images/2-73.gif)
(commands used to generate figure above)
Let's look at an example. Let's find the gradient of
g
(
x,y
) =
at (3,1), the tangent line to the level curve with level 9, and a vector
m
which is tangent to the level curve at (3,1).
Solution:
To begin with, the gradient of
g
(
x,y
) is grad(
g
) =
, so that at (3,1) we have grad(g) =
. Moreover, the level curve of
g
(
x,y
) with
k=
9 is
, which leads to
Thus,
y'
=
, and at
x=
3 we have
y'
(3) =
. Thus, a run of 3 leads to a rise of -2, so that
m
=
is tangent to the curve at (3,1). However,
grad(g) . m = 3(6) - 2(9) = 0
Thus, the gradient and the tangent vector m are orthogonal, as is shown in the plots below:
>
p1a:=arrow([3,1],vector([6,9]),0.1,0.4,0.1,color=green):
p1b:=arrow([3,1],vector([3,-2]),0.1,0.4,0.1,color=red):
p2:=plot(9/x^2,x=1..10,color=black,thickness=2):
p3:=plot(1-2*(x-3)/3,x=0..10,color=red):
p4a:=textplot([7,6,"grad(g)(3,1)"],align={ABOVE,RIGHT},font=[TIMES,ROMAN,12]):
p4b:=textplot([5,-0.51,"m"],align={BELOW,LEFT},font=[TIMES,BOLD,12]):
display([p1a,p1b,p2,p3,p4a,p4b],view=[1..10,-3..10],scaling=constrained);
>